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7.2. Improper integrals: integrand unbounded

There is a second type of improper integral, in which the function is unbounded in the interval [a, b] If the function f(x) becomes arbitrarily large in the interval [a,b], then it is possible to have arbitrarily large approximating

sums f ( i ) xi no matter how fine the partition may be by choosing a 1

i 1

that makes f( 1) large.

Let f(x) be continuous at every number in [a, b] except a, and become

arbitrarily large for values in (a, b). If lim f (x)dx exists, the function f(x)

0 a

is said to have a convergent improper integral from a to b. The value of the

		b			b
limit is denoted			f (x)dx . If lim			f (x)dx does not exist, the function f(x) is
				0
		a			a
said to have a divergent improper integral from a to b; in brief b							f (x)dx is
not defined.						a
not defined.
	In a similar manner, if f(x) is unbounded only near b (Fig. 2.3), define
b		b
	f (x)dx as lim			f (x)dx .
	0
a			a

It may happen that a function behaves well everywhere in the interval [a, b] except at the number c, distinct from a to b (Fig. 2.4), where it may be

у=f(x)

О		а	b-ε b х

Fig. 2.3

infinite. In that case b f (x)dx

у=f(x)

О а c-cε c+δ b х

Fig. 2.4

makes no sense. In such a case consider

162

c	f (x)dx	and b	f (x)dx . If both exist, then the integral				b	f (x)dx is said to be
a		c					a
convergent and have the value c				f (x)dx	+ b	f (x)dx . More generally, if a
			a		c

function f(x) has an infinite range of values as well as a point where it becomes infinite, break the entire integral into the sum of the integrals each of which has only one of the two basic “troubles”, either an infinite range or an endpoint where the function is infinite. For instance, the improper integral

	1	dx is troublesome for four reasons:	lim	1	,	lim	1	, and
x4			x 0 0	x4		x 0 0	x4

the range extends infinitely to the left and also to the right. To treat the integral, write it as the sum of four improper integrals of the two basic types:

	1	0	1
dxx4	dxx4 dxx4 dxx4 dxx4 .
		1	0	1

All four of the integrals on the right have to be convergent for

be convergent. As a matter of fact, only the first and last are. So

dx to

dx is

divergent.

Just as substitution in a definite integral is valid as long as the same substitution is applied to the limits of integration, substitution in improper integrals is also permissible.

T 7. Typical problems

In exercises 1 and 2 determine whether the improper integrals are convergent or divergent. Evaluate the convergent ones.

		dx
1.		dx		.
1.	1	x	2	.
1	1	x
1
Solution. By definition
					dx			A	dx			A
								A				A
							lim				lim arctg x
					1 x	2	lim		1 x	2	lim arctg x
			1		1 x		A	1			A	1
			1					1				1

163

lim (arctg A arctg1)

Therefore,

converges to

and the area in question is

1 x2

2. dx .

Solution.

lim

dx lim ln

lim ln

=¥ .

1 x

A ¥

Therefore, the given integral is divergent.

3. Examine the improper integral

for all

p .

Solution. By example 2 if

p 1 the integral is divergent. If

p 1 , then

A dx

x p 1

A p 1

lim

p 1

1 x

A 1

A p 1

p 1,

p 1

p 1.

In exercises 4–5 determine whether the improper integrals are convergent or divergent. Evaluate the convergent ones.

		dx
4.			.
	2	6x 10
x

Solution. The substitution x – 3 = t transforms the integral to

t2 1 .

By definition

	dt	a	dt		dt		a	dt		B	dt
						lim			lim			,
	2		2		2			2			2
t	1	t	1	a t	1	A A t		1	B a t		1

where a is any number. Thus:

164

lim ò

A -¥ A

Next consider

Hence

= lim

arctg t

a = arctg a - lim arctg A =

t2 +1

A -¥

arctg a + p ;

lim

= lim arctg t

-arctg a .

B a t

B ¥

, namely

is convergent and equals .

6x 10

ln x

dx .

Solution.

ln x

lim

ln xd (ln x)

lim

2 x

¥ .

Therefore, the given integral is divergent.

In Exercises 6 to 8 examine the improper integral.

cos3 x dx .

Solution. The integrand

cos x

is the alternative function if x 1.

As is

generally

known

cos x

then

cos x

. By Exercise 3 the

dx3

integral

is convergent ( p 3 1 ), so is the given integral.

ln(1 x2 ) 2 ln x dx .

Solution. Write the integral in the form

x 2

dx ln 1

dx .

165

																	dx
By Exercise 3 the integral																	dx			is convergent ( p 2 1). Recall also
By Exercise 3 the integral																	2			is convergent ( p 2 1). Recall also
																1	x
																1
								1
		ln		1
		ln		1				x	2
that lim								x				1 , so is the given integral.
that lim						1						1 , so is the given integral.
x						1
					x2
2				dx
8.				dx							.
8.	x		2	2x						3	.
1	x			2x						3
Solution. At x = 1 the integrand is undefined, and near x = 1 it is
unbounded. It is necessary to examine the improper integral
				2						dx						2				dx						2				d(x 1)
										dx					lim					dx			lim							d(x 1)
						x	2		2x				3		lim			x	2	2x 3			lim						(x 1)			2	4
				1		x			2x				3		0	1		x		2x 3				0	1				(x 1)				4
				1												1									1
							lim						1	ln	x 1	2				1	lim(ln		1	-ln				e			) =¥.


															x 3
									0				4			1				4 e0			5			4	+e
													4			1				4 e0			5			4	+e
Therefore, the given integral is divergent.
																				1	dx
9. Examine the improper integral																					dx	for all .
9. Examine the improper integral																						for all .
																				0	x
																				0

Solution. At x = 0 the integrand is undefined, and near x = 0 it is unbounded. It is necessary to examine the two improper integrals. If 1 , then

1	dx				1	dx				x1		1			1	1
1	dx				1	dx				x1		1			1	1
		lim					lim						= lim
	x	0				x	0			1					1
0	x					x								0	1	1
0													1	0		1
				1					1				1
		=				lim								, if 1,
														, if 1,

			1					1
			1				o	1			1				1.
													, if		1.

Next consider for 1

1			1				1

dx			dx
		lim			lim ln	x	.

0	x	0		x	0

166

	1	dx
Conclusion. The second type of improper integral		dx	is convergent if
Conclusion. The second type of improper integral			is convergent if
	0	x
1 and is divergent if	1 .

T 7. Exercises for class and homework

Determine whether the improper integrals are convergent or divergent. Evaluate the convergent ones.

					dx																dx
1.					dx			.									2.				dx			.			3.			e 3x dx .				4.
1.					6			.									2.							.			3.			e 3x dx .				4.
			1		x														1	5	x 2									0
				xdx
				xdx					.
				2					.
			1 x
									dx													dx
5.									dx							.	6.					dx						. 7.		xe x2 dx .				8.
5.						x	2									.	6.			(1 x)x						2		. 7.		xe x2 dx .				8.
						x			2x 10										1	(1 x)x										0
																			1											0
							x dx .										1			dx							2				dx			2	xdx
		arctg2														9.				dx					.		10.				dx		.	11.	xdx		.
		arctg2														9.									.		10.			2	4x 3		.	11.			.
1				x													0		1 x 2									0 x			4x 3			1	x	1
				1
12. e								dx					.
12. e								2					.
				0		x ln					x
				0
Determine whether the improper integrals are convergent or divergent
							xdx											x	3 1
13.							xdx					.			14.			x	3 1			dx .						15. ln(1 x 2 )dx .						16.
13.						1	3					.			14.				4			dx .						15. ln(1 x 2 )dx .						16.
				0		1	x										1		x								1					x
			dx
			dx							.
			x ln ln x							.
e	2
e
				1					xdx								1		dx								1			dx
17.									xdx					.	18.				dx				.					19.		dx		.		20.
17.														.	18.								.					19.	sin x			.		20.
				0			1 x 4										0	e	x		1						0		sin x
1					dx
					dx						.
					3						.
0				x					x

167

Answers

1. 15 . 2. Diverges. 3. 13 . 4. Diverges. 5. 3 . 6. 1 ln 2 . 7. 12 . 8. 4 12 ln 2 .

9. 2 . 10. Diverges. 11. 83 . 12. 1. 13. Converges. 14. Diverges. 15. Diverges. 16. Diverges.

17. Converges. 18. Converges. 19. Diverges. 20. Converges.

T 7. Individual test problems

7.1. Determine whether the improper integrals are convergent or divergent. Evaluate the convergent ones.

				xdx																16xdx																									x3dx
7.1.1.										.						7.1.2.													.								7.1.3.															.
7.1.1.							4			.						7.1.2.								4					.								7.1.3.															.
0 16x									1									1 16x								1													0					16x4 1
				xdx												0									xdx																			x2 dx
7.1.4.				xdx								.				7.1.5.									xdx						.						7.1.6.							x2 dx							.
1	16x4 1																					(x2 4)3															0			3 (x3 8)4
						xdx																			xdx																				dx
7.1.7.						xdx							.			7.1.8.									xdx							.				7.1.9.									dx								.
7.1.7.													.			7.1.8.																.				7.1.9.			(x					2	4x 5)								.
0 4 (x2 16)5																4					x2 4x 1																	1	(x						4x 5)
							xdx																				16dx												arctg 2xdx
7.1.10.														.	7.1.11.																					. 7.1.12.											2					.
7.1.10.	x		2			4x 5								.	7.1.11.							(4x					2		4x 5)							. 7.1.12.						(4x					2	1)				.
1	x					4x 5												1/ 2				(4x							4x 5)										0			(4x						1)
1																		1/ 2																					0
								xdx																(x 2)dx																		3 x					2
7.1.13.															.	7.1.14.																			.		7.1.15.											dx .
7.1.13.					2										.	7.1.14.																			.		7.1.15.					x		2				dx .
0 4x								4x				5							0 3 (x2 4x 1)4																				0			x					4
								4dx															2				arctg x
7.1.16.														.		7.1.17.											arctg x						dx .				7.1.18. x sin xdx .

		x(1 ln								2		x)											4x						2		1
1		x(1 ln										x)						0					4x								1								0
1																		0																					0
-1						7dx																						dx																			dx
7.1.19. ò						7dx						.			7.1.20.													dx									. 7.1.21.										dx					.
7.1.19. ò				x		2						.			7.1.20.					(9x					2	1) arctg								2	3x		. 7.1.21.							x	2	(x 1)						.
-¥				x				-4x								1/ 3				(9x						1) arctg									3x						1			x		(x 1)
-¥																1/ 3																									1
0							x2								x																	dx															3x
7.1.22.																	dx . 7.1.23.																					.7.1.24. xe											dx .
7.1.22.							3									2	dx . 7.1.23.													2								.7.1.24. xe											dx .
x									1			1		x												1	(6x				5x 1)												0
¥				dx																				dx																				dx
7.1.25. ò				dx							.				7.1.26.									dx							.				7.1. 27.									dx						.
7.1.25. ò		x		2							.				7.1.26.				2x		2		2x 1								.				7.1. 27.				x		2		3x						2	.
1		x					+2x									0			2x				2x 1														3		x				3x						2
1																0																					3

168

	dx					dx				dx
7.1.28.	dx		.7.1.29.			dx	.7.1.30.			dx			.
7.1.28.	x(ln x 1)	2	.7.1.29.	9x	2	9x 2	.7.1.30.		2			x	.
e2	x(ln x 1)		1	9x		9x 2	2	(x	2	4)	arctg	x
								(x		4)	arctg	2

7.2. Determine whether the improper integrals are convergent or divergent. Evaluate the convergent ones.

1/ 3 e3 1/ x																	3				dx									1			dx
7.2.1.												dx .					7.2.2.									.			7.2.3.									.
7.2.1.						x		2				dx .					7.2.2.									.			7.2.3.		3							.
	0					x											1	x	2 6x 9											0		2		4x
	1							dx									1													3					dx
7.2.4.								dx					.				7.2.5. ln(3x 1) dx .												7.2.6.						dx								.
7.2.4.							2						.				7.2.5. ln(3x 1) dx .												7.2.6.														.
1/ 4 20x								9x 1									1/ 3 3x 1													1 3 (3 x)5
	1							ln 2									2 / 3	3	ln(2 3x)											1		xdx
7.2.7.																dx .	7.2.8.									dx		. 7.2.9.								.
7.2.7.										2						dx .	7.2.8.				2 3x					dx		. 7.2.9.				x			4	.
1/ 2		(1 x) ln									(1 x)						0				2 3x									0 1		x
	/ 6						cos 3xdx										1		2xdx											0				dx
7.2.10.							cos 3xdx									.	7.2.11.		2xdx					.				7.2.12.						dx					.
7.2.10.																.	7.2.11.							.				7.2.12.				3							.
	0			6 (1 sin 3x)5													0		1 x4										1/ 3				1 3x
	1	2e				1		2	arcsin x								/ 2				e tg x									1				dx
7.2.13.		2e											dx .				7.2.14.				e tg x				dx .			7.2.15.						dx							.
7.2.13.													dx .				7.2.14.			cos 2x					dx .			7.2.15.				5	3 4x								.
	0						1 x2										0			cos 2x								3 / 4					3 4x
	2							dx												sin xdx										0					dx
7.2.16.								dx							.		7.2.17.			sin xdx						.		7.2.18.							dx									.
7.2.16.		5 4x x2													.		7.2.17.									.		7.2.18.																.
	1	5 4x x2										4					/ 2		7 cos2						x					3 / 4				4x 3
	2							xdx									1/ 3					dx								3		3 9xdx
7.2.19.								xdx								.	7.2.20.					dx						. 7.2.21.				3 9xdx							.
7.2.19.																.	7.2.20.	9x			2	9x 2						. 7.2.21.											.
	1			(x2 1)3 ln 2													0	9x				9x 2								0 3 9 x2
	/ 2				3sin 3 xdx .												1	x4 dx										2				x2 dx
7.2.22.																	7.2.23.	x4 dx					.						7.2.24.			x2 dx								.
7.2.22.																	7.2.23.						.						7.2.24.											.
	0							cos x									0 3 1 x5											0				64 x6
	1						dx										5				x2 dx									1/ 4					dx
7.2.25.							dx				.						7.2.26.				x2 dx						.	7.2.27.							dx							.
7.2.25.			9								.						7.2.26.										.	7.2.27.					3									.
	1/ 2				1		2x										1	31(x3 1`)												0				1 4x
	4					10xdx											3 / 2					dx								1/ 2					dx
7.2.28.						10xdx								.		7.2.29.						dx							. 7.2.30.						dx										.
7.2.28.														.		7.2.29.													. 7.2.30.				(2x				1)				2				.
	0	4 (16 x2 `)3															1			3x x2						2				0			(2x				1)

169

Topic 8. Application of the definite integral

Finding the area of a region. How to find the arc length. Computing volume by parallel cross sections. Computing the volume of a solid of revolution. The centroid of a plane region. Work.

Literature: [1, section 9], [2, section 2, ch. 2.2], [4, section 7, §24], [6, section 10], [7, section 12], [9, §41].

T8. Main concepts

8.1. Computing areas

8.1.1. Area in rectangular coordinates

Consider the area of the region bounded by y f (x) ( f (x) 0 ) and the x-

axis from x = a to x = b, as shown in Fig.2.5. The area S of the entire region is found by

S f (x)dx.

However, if a function changes the sign finitely in [a; b] (Fig. 2.6) the area is found according to the following formula

S f (x)dx .

y=f(x)

b x

Fig. 2.5

Fig. 2.6

Let f1 and f2 be two continuous functions such that

f1 (x) f 2 (x) for all

x in the interval [a, b]. Let S be the region between the curve y f1 (x) and the curve y f 2 (x) for x in [a, b], as shown in Fig. 2.7. The area of S is then given by

170

S [ f2 (x) f1(x)]dx

It should be obvious to you that knowing the points of intersection is important in determining the limits of integration. For instance, the sketch of the region appears in Fig. 2.8. To find where the curves intersect, we solve the

system of equations y f1 (x)

= x1 and b = x2.

y y=f2(x)

y=f1(x)

Оa

Fig. 2.7

and y f 2 (x) . In this case, if x1

x2, then a

y y=f2(x)

y=f1(x)

b x

О a

b x

Fig. 2.8

Sometimes area can more easily be determined by summing areas of horizontal elements rather than vertical elements.

Let S be the region between the curve x g1 ( y) and the curve x g 2 ( y) for y in [c, d], as shown in Fig. 2.9 and 2.10. The area of S is then given by

			d
		S [g2 ( y) g1 ( y)]dy.
			c
y					y
d					d	x=g2(x
d					d

c x=g1(x	S		x=g2(x		x=g1(x	S
			x=g2(x
					c	x
					c
О				x	О

Fig. 2.9	Fig. 2.10
8.1.2. Area, if an outline is given parametrically
The path of some particle is	given parametrically by the

equations x x(t) , y y(t) 0 , t [ ; ] . The area of S is then given by

Sy(t)dx(t) y(t)x (t)dt,

171

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